Question

A man has 3 friends. If N is the number of ways he can invite one friend every day for dinner on 6 successive nights so that no friend is invited more than 3 times then the value of $ \dfrac{N}{170} $ is.

Answer 1

Hint: We will first start by finding the different cases in which we can group the number of times a friend can be invited in 6 days. Then we will find the actual number of ways in which we can group different cases and then find the total ways.

Complete step-by-step answer:
Now, we first let the three friends of the man be,
A, B, C
Now, it has been given to us that a friend cannot be invited more than 3 times. So, we have the different groups in which we can fill the 6 nights of the man as,
Case I (3, 3, 0) …………(1)
Case II (3, 2, 1) …………(2)
Case III (2, 2, 2) …………(3)
Now, we have the different ways the man can have all his night filled with his friend. Now, we have to find the actual number of permutations for each case by taking in account the different friends that will have dinner with the man. For example, if A have dinner 3 times or B or C. So, we have,
From case I we have the number of ways of grouping (3, 3, 0) in 6 nights as $ \dfrac{6!}{3!\times 3!\times 2!} $ by using the formula that number of ways of grouping $ a+b+c $ in (a, b, c) is $ \dfrac{{{\left( a+b+c \right)}^{2}}!}{a!\times b!\times c!} $ .
Now, we can also permute the three friends in group as B. So, total ways for case I is,
 $ \dfrac{6!}{3!\times 3!\times 2!}\times 3! $
Now, similarly for case II, we have (3, 2, 1).
Total ways will be $ \dfrac{6!}{3!\times 2!\times 1!}\times 3! $ .
Now, for case III we have (2, 2, 2).
Total ways will be \[\dfrac{6!}{2!\times 2!\times 2!\times 3!}\times 3!\].
So, for total ways in all three cases we have,
 $ \begin{align}
  & \dfrac{6!\times 3!}{3!\times 3!\times 2!}+\dfrac{6!\times 3!}{3!\times 2!\times 1!}+\dfrac{6!\times 3!}{2!\times 2!\times 2!\times 3!} \\
 & \Rightarrow \dfrac{4\times 5\times 6}{2}+3\times 4\times 5\times 6+\dfrac{3\times 4\times 5\times 6}{2\times 2} \\
 & \Rightarrow 10\times 6+18\times 20+3\times 30 \\
 & \Rightarrow 60+360+90 \\
 & \Rightarrow 150+360 \\
 & \Rightarrow 510 \\
\end{align} $
So, the value of N is 510.
 $ \begin{align}
  & \Rightarrow \dfrac{N}{170}=\dfrac{510}{170} \\
 & \Rightarrow \dfrac{N}{170}=3 \\
\end{align} $

Note: It is important to note that we have used the concept that if we have n different objects and we have to divide them into r groups of sizes $ {{a}_{1}},{{a}_{2}},{{a}_{3}},......{{a}_{r}} $ the number of ways to do so will be $ \dfrac{n!}{{{a}_{1}}!\times {{a}_{2}}!\times ......{{a}_{r}}!} $ .